Lemma 2 [1]}, [2]}
Let \(\Xi =(a_{js})_{N\times N}\)  be a \(N\times N\)  matrix with
column vectors \(\lbrace \alpha _j\rbrace \) . \(\Gamma =(\gamma _{js})_{N\times N}\)  is an \(N\times N\) 
operator matrix and each \(\gamma _{js}\)  is a operator. Then we have
\(\sum ^N_{j=1}|\Gamma _{j}\ast \Xi |=\sum ^N_{j=1}|(\Gamma ^{T})_{j}\ast \Xi ^{T}|,\) 
